P1=1 T1=17=2900K P2=32 bar
P1P2=T1T2k−1k
T2=T1(P1P2)kk−1
T2=290×1320.2857=780.580K
P4=5P2=532=6.4 bar
T4=T1(P1P4)kk−1
T4=290×16.40.2857=483.850K
Heat addition:
Qin=mCv△T=1×0.717×(T2−T1)=1×0.717×(780.58−290)=351.74kgKkJ
Heat rejection :
Qout=mCv△T=1×0.717×(T4−T1)=1×0.717×(483.85−290)=193.85kgKkJ
Efficiency of the cycle:
ηth=QinQin−Qout=351.74351.74−193.85=0.4488=45%
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